Pratical Lecture 4
Mecânica de Fluidos Ambiental 2015/2016
Resolution
1
1
1
V12  p2  V22  pa  V22
2
2
2
V1 D12  V2 D22
p1 
p1  gh  pa  0
1
1
1
2
2
pa  gh  V1  p2  V2  pa  V22
2
2
2
2
2
2
2
2  D1 
V1 D1  V2 D2  V2  V1  2 
 D2 
D
1
1
V12  gh  V12 
2
2
D
2
1
2
2
V1 
2 gh
 D12 
1   2 
 D2 
2



2
2
Resolution
1
V32  gL  p1
2
1
p2  V22  gH  p1
2
V2  V3  V
p3 
p2  p1  pa
1
V22  gH  0
2
V2  2 gH
1
V32  gL  p1  0
2
1
p3   V32  gL
2
1
p3    2 gH   gL   g H  L 
2
p3 
Resolution
1
1
V12  gz1  V22  gz 2
2
2
1 2
V2  V12  g( z1  z 2 )
2
1  Q2 Q2 

 2   g( z1  z 2 )
2  A22
A1 

Q 

2g( z1  z 2 )

1
1
 2
A22
A1
2 * 9.8 * 0.1
 3.8m2 / s
1
1
 2
2
1.6
2
V2  2.35m / s
V1  1.9m / s
With a new iteration the
calculation can be improved
QE  QS  Qup
  
QE   v .n dA  U 0b
0
 y  y 3 
3   

    

U0b  3 y 2 1 y 4 
QS   U 0 


bdy 
3 
0
2
2
2

4



 0






U  3   5U 0b
QS  0 b 
 
2  2
4
8
3U 0b
Qup 
8
VS  VE
VE  V0
 0Q V0   V0   F
F  2  0QV0
Resolution



 
 p



ui dV    ui u.n    ui .n dA      gi dVol

t vc
 xi

surface




u
u
 i .n dA
F  b
surface
Exit Surface

U U





u
u
.
n
dA

b
2

 i
 
L
 S  b
surface
0
L
2
y

2 L
2
y
y 
U  
U  7
 S  b2     y 
   b2    L
L 3L  0
2 
2 3
2
2
3
2
2
2
2
2
  U 2

U
y
U
y







dy  b 2      2 
      dy
 2 
 2  L  2   L  
0
L
Momentum flux at the entrance
• The fluid leaving through the area of lower momentum entered through
an surface shorter than 2L. Using the mass conservation one gets:
m 
 u dA
i
surface
U U y 
 m  2b   
dy 
0
2
2
L


L
U
 m  2b
2
bLEU 
LE 
L

y2 
U 3
3
y


2
b

L

bLU


2L  0
2 2
2

3
bLU
2
3
L
2
• The Momentum entering is:
 E  bU 2
3
L
2
And the force and Cd
2
3
1
U  7
7 3
F  b 2    L  bU 2 L  bU 2 L     bU 2 L
2
3
2 3
6 2
1
bU 2 L
F
1
3
CD 


bU 2 L
bU 2 L
3
• Let us consider a large number of plates, as in a
real turbine.
Resolution
• In this case there is movement of the plate. The linear speed is:
Vc  R
• The relative velocity has the same modulus and opposite sign. The
discharge is the jet discharge, if we assume many plates.
VS r  VE r
VE r  V j  R 
VS r  V j  R 
Q  Vj
D 2
Q j  V j  R   V j  R   F
F  2 Q j V j  R 
4
Pot  2 Q j V j  R R
• Maximum power is obtained when the derivative
in relation to the angular velocity is null:
Pot  2 Q j V j  R R
 dPot 

  2 Q j V j  R R  R 2 Q j  R   0
 d 
V j  R   R
V j  2R
• In this case
VS r  V j  R   R
Vabsolute S  R  VS r  R  R  0
• Whole jet the kinetic energy would be extracted. Efficiency would
be 100%.
• Mass balance
• Momentum Balance
• Energy Balance
Resolution
UA1  UA2  Q
 U 2 Q   U1 Q  PA1  F
1
1


2
2
 P  U    P  U 
2
2

1 
2
2


1
1
A
2
2
2
2
P1   U 2  U1  U 2 1  2 
2
2
 A1 


Resolution
 U 2 Q   U 1 Q  PA1  PA2  F
 U Q  UUh
1
gh 2
2
1
F   g h12  h22  U 22 h2  U 12 h1
2
P


• Bernoulli
1
1
U 22  gh2  U 12  gh1
2
2
• Mass conservation
U 2 h1  U 1h1
h
1
1
U 22  gh2  U 22  2
2
2
 h1
U2 
2 g h1  h2 
 h2
1 
h
 1




2
2


  gh1

Resolution considering U  U
1

2

1
F   g h12  h22  2 g h1  h2 h2
2
This equation states that the force must balance the hydrostatic pressure plus the
momentum exiting from under the sluice gate. The force tends to zero when h2
goes to h1 . The force would grow linearly to be maximum for h2 =0
dF
 gh2  2 g  h2  h1  h2   2 g h1  h2 
dh2
If the momentum entering into the system was relevant one had to subtract it
from the force:
F 


1
g h12  h22  U 22 h2  U 12 h1
2
 h2
1
F   g h12  h22  U 22 h2  U 22 h1 
h
2
 1

F 

1
g h12  h22
2






2





h 2 g h1  h2  
 h2 1  2
2 
h1
 h2  

1 

h 
 
 1 

Description of the problem
Vin
V
Vj
V
•
•
Vj is the jet velocity relative to the boat and Vin is the inlet velocity also relative to the boat.
Qj is the jet discharge (equal to the inlet discharge).
The boat “sees” the water entering into the control volume at its own velocity and leaving it
at the relative velocity (Vj).
F  Q j V j  V   kV 2
kV 2  Q jV  Q jV j  0
V
 Q j 
Q 
2
j
 4kQ jV j
2k
The boat velocity depends on the jet velocity, on the discharge and on friction.
If there was no friction there would be no force and the maximum boat speed
would be the jet speed.
• If we had considered a control volume without the green area, we
should consider Vin instead of V, but in that case we should have
computed the pressure force in front of the boat, computed as:
Q j Vin  V   Fin
•
•
This is the force applied over the fluid inside the green volume. The pressure
reduction at the interface between both is the symmetrical of that force.
This force must appear in the application to the blue control volume:
Q j V j  Vin   F  (Fin )
Q j V j  Vin   F  Q j Vin  V 
F  Q j V j  Vin  Vin  V   Q j V j  V 
• Which is the same result as before.
Vin
V
Vj